Finding Tangents and Normals Revision Notes | GCSE AQA Higher Further Maths

1. Why Tangents and Normals Matter

Picture a ball rolling along a curved ramp. At any instant, the direction it travels is along the tangent to the curve at that point — not along the curve itself. That single straight line captures the curve's behaviour at one precise moment.

The normal is the line at right angles to the tangent at the same point. Together, these two lines appear regularly in GCSE exam questions and require a clear three-step method: differentiate, find the gradient, then build the line equation.

2. The Key Formulae

The point-gradient form $y - y_1 = m(x - x_1)$ is the engine behind both equations. You already know how to use it for straight lines — here you are feeding it a gradient produced by differentiation rather than one read from a diagram.

3. The Three-Step Method

Every tangent-and-normal question follows the same three steps, regardless of how the curve looks. Nail these steps and the question becomes routine.

• Step 1 — Differentiate: find $\frac{dy}{dx}$ for the curve.
• Step 2 — Substitute: put the given $x$-value into $\frac{dy}{dx}$ to get the gradient $m$ at that point.
• Step 3 — Build the line: use $y - y_1 = m(x - x_1)$ for the tangent, or $y - y_1 = -\frac{1}{m}(x - x_1)$ for the normal.

If the question gives only an $x$-value, substitute it into the original equation of the curve to find the $y$-coordinate before moving to Step 3. Missing this step is one of the most common sources of lost marks.

4. Worked Example — Finding the Tangent

Worked example: Finding the Equation of the Tangent

The curve $C$ has equation $y = x^3 - 4x + 1$. Find the equation of the tangent to $C$ at the point where $x = 2$.

1. $y = (2)^3 - 4(2) + 1 = 8 - 8 + 1 = 1$ Point of contact: $(2, 1)$ — Substitute $x = 2$ into the original curve equation. The point $(2, 1)$ is where the tangent touches the curve. [M1]
2. $\frac{dy}{dx} = 3x^2 - 4$ — Differentiate term by term: $x^3 \to 3x^2$, $-4x \to -4$, constant $1 \to 0$. [M1]
3. $m = 3(2)^2 - 4 = 3 \times 4 - 4 = 12 - 4 = 8$ — Substitute $x = 2$ into $\frac{dy}{dx}$. The gradient of the tangent at $(2, 1)$ is $8$. [A1]
4. $y - 1 = 8(x - 2)$ $y - 1 = 8x - 16$ $y = 8x - 15$ — Use $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, 1)$ and $m = 8$. Rearrange to $y = mx + c$ form. [A1]

Answer: $y = 8x - 15$

5. Worked Example — Finding the Normal

Once you have the tangent gradient, the normal gradient follows immediately from the negative reciprocal rule. The point of contact is identical — only the gradient changes.

Worked example: Finding the Equation of the Normal

The curve $C$ has equation $y = x^3 - 4x + 1$. Find the equation of the normal to $C$ at the point where $x = 2$. Give your answer in the form $ay + bx = c$ where $a$, $b$, and $c$ are integers.

1. Point: $(2, 1)$, tangent gradient: $m = 8$ — These were found in the previous example. No need to repeat the differentiation from scratch.
2. $m_{\text{normal}} = -\frac{1}{8}$ — The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $8$. [M1]
3. $y - 1 = -\frac{1}{8}(x - 2)$ $8(y - 1) = -(x - 2)$ $8y - 8 = -x + 2$ $8y + x = 10$ — Use $y - y_1 = m_{\text{normal}}(x - x_1)$ with $(2, 1)$ and $m = -\frac{1}{8}$. Multiply through by $8$ to clear the fraction, then rearrange. [A1]

Answer: $8y + x = 10$

The graph shows the curve $y = x^3 - 4x + 1$ (blue), the tangent $y = 8x - 15$ (steeper line touching at $(2,1)$), and the normal $y = -\frac{1}{8}x + \frac{5}{4}$ (shallow line through $(2,1)$). Notice how the tangent and normal cross at exactly the point of contact and are clearly perpendicular to each other.

6. Guided Practice

The method is consistent across all curve types at GCSE. Work through these problems using the three-step approach: differentiate, substitute, build the line.

7. Exam Technique and Common Mistakes

Examiners award marks at each stage of the method, so a clear, structured layout protects your marks even if arithmetic slips occur. Write $\frac{dy}{dx} =...$, then $m =...$, then the line equation — each on its own line.

A final check worth building into your routine: substitute the point of contact into your tangent or normal equation. If the left-hand side equals the right-hand side, the equation is correct. This takes ten seconds and can save several marks.

8. Exam-Style Question

This question mirrors the style and demand of a GCSE exam question on this topic. Work through it using the three-step method before checking the solution.

Worked example: Exam-Style: Tangent and Normal at the Same Point

The curve $C$ has equation $y = 2x^2 - 3x + 5$. The point $P$ lies on $C$ where $x = 2$. Find the equation of (a) the tangent to $C$ at $P$, and (b) the normal to $C$ at $P$.

1. $y = 2(2)^2 - 3(2) + 5 = 2 \times 4 - 6 + 5 = 8 - 6 + 5 = 7$ Point $P = (2, 7)$ — Substitute $x = 2$ into the curve equation. $2 \times 4 = 8$, then $8 - 6 + 5 = 7$. [M1]
2. $\frac{dy}{dx} = 4x - 3$ — Differentiate term by term: $2x^2 \to 4x$, $-3x \to -3$, $5 \to 0$. [M1]
3. $m = 4(2) - 3 = 8 - 3 = 5$ — Substitute $x = 2$ into $\frac{dy}{dx}$. The tangent gradient is $5$. [A1]
4. $y - 7 = 5(x - 2)$ $y - 7 = 5x - 10$ $y = 5x - 3$ — Point-gradient form with $(2, 7)$ and $m = 5$. Rearrange to $y = mx + c$ form. [A1]
5. $m_{\text{normal}} = -\frac{1}{5}$ — Negative reciprocal of $5$ is $-\frac{1}{5}$. [M1]
6. $y - 7 = -\frac{1}{5}(x - 2)$ $5(y - 7) = -(x - 2)$ $5y - 35 = -x + 2$ $5y + x = 37$ — Use the same point $(2, 7)$ with $m = -\frac{1}{5}$. Multiply through by $5$ to clear the fraction. [A1]

Answer: (a) Tangent: $y = 5x - 3$ (b) Normal: $5y + x = 37$

Every tangent-and-normal question at GCSE reduces to the same three-step method: differentiate the curve, substitute the $x$-value to get the gradient, then use the point-gradient form. Mastering this method covers the full specification requirement for finding the equation of a tangent and normal at any point on a curve.